Question

[o0o nghịch ngợm o0o]

1, Cho \(\frac{a}{2016}=\frac{b}{2017}=\frac{c}{2018}\). Tính M= 4* [ a-b] * [ b-c]- [c-a]²

2, Cho \(\frac{2a+b+c}{b+c}=\frac{2b+c+a}{c+a}=\frac{2c+a+b}{a+b}\) biết a+b+c khác 0. Tính M=\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

3, TÍnh giá trị biểu thức : M= \(\frac{y^2}{\left[z+t+x\right]^2}+\frac{z^2}{\left[t+x+y\right]^2}+\frac{t^2}{\left[x+y+z\right]^2}+\frac{x^2}{\left[y+z+t\right]^2}\)

Nhờ mn giúp đỡ, mk đang gấp 

Answer 1

[vũ tiền châu]

Ta có \(\frac{2a+b+c}{b+c}=\frac{2b+c+a}{c+a}=\frac{2c+a+b}{a+b}\Rightarrow\frac{2a}{b+c}+1=\frac{2b}{a+c}+1=\frac{2c}{a+b}+1\)

=> \(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\Rightarrow\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\frac{3}{2}\)

^_^ 

Answer 2

[tth_new]

Bài 1: Đặt \(\frac{a}{2016}=\frac{b}{2017}=\frac{c}{2018}=k\)

\(\Rightarrow\hept{\begin{cases}a=2016k\\b=2017k\\c=2018k\end{cases}}\).Thay vào M,ta có:

 \(M=4\left(2016k-2017k\right)\left(2017k-2018k\right)-\left(2018k-2016k\right)^2\)

\(=4.\left(-1k\right)\left(-1k\right)-\left(2k\right)^2\)

\(=4k^2-4k^2=0\)